Question:
Three identical liquid drops, each carrying the same charge, coalesce to form a single drop. The ratio of the potential of the large drop and the smaller drop is \( 3^{1/3} \). Then, \( N \) is:
Three identical liquid drops, each carrying the same charge, coalesce to form a single drop. The ratio of the potential of the large drop and the smaller drop is \( 3^{1/3} \). Then, \( N \) is:
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When drops coalesce, the volume adds up and the radius increases. The potential of the drop is directly proportional to the radius.
Updated On: Jan 29, 2026
- A
- B
- C
- D
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The Correct Option is B
Solution and Explanation
Step 1: Understand the relation for potential.
The potential \( V \) of a spherical drop is given by the formula: \[ V = \frac{4 \pi \varepsilon_0 R q}{R}, \] where \( R \) is the radius and \( q \) is the charge. The total charge \( Q \) is the same for all drops. Step 2: Calculate the potential.
When the drops coalesce, their volume adds up, so the radius of the large drop becomes \( R_{\text{large}} = 3^{1/3} R_{\text{small}} \). The potential is proportional to the radius of the drop. Therefore, the ratio of the potentials is \( 3^{1/3} \), and \( N = 3 \). Final Answer: \[ \boxed{3}. \]
The potential \( V \) of a spherical drop is given by the formula: \[ V = \frac{4 \pi \varepsilon_0 R q}{R}, \] where \( R \) is the radius and \( q \) is the charge. The total charge \( Q \) is the same for all drops. Step 2: Calculate the potential.
When the drops coalesce, their volume adds up, so the radius of the large drop becomes \( R_{\text{large}} = 3^{1/3} R_{\text{small}} \). The potential is proportional to the radius of the drop. Therefore, the ratio of the potentials is \( 3^{1/3} \), and \( N = 3 \). Final Answer: \[ \boxed{3}. \]
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